{"id":546,"date":"2021-02-04T16:10:28","date_gmt":"2021-02-04T16:10:28","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=546"},"modified":"2022-03-16T22:19:23","modified_gmt":"2022-03-16T22:19:23","slug":"fundamental-theorem-of-calculus","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/fundamental-theorem-of-calculus\/","title":{"raw":"Fundamental Theorem of Calculus","rendered":"Fundamental Theorem of Calculus"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>State the meaning of the Fundamental Theorem of Calculus, Part 1<\/li>\r\n \t<li>Use the Fundamental Theorem of Calculus, Part 1, to evaluate derivatives of integrals<\/li>\r\n \t<li>State the meaning of the Fundamental Theorem of Calculus, Part 2<\/li>\r\n \t<li>Use the Fundamental Theorem of Calculus, Part 2, to evaluate definite integrals<\/li>\r\n \t<li>Explain the relationship between differentiation and integration<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Fundamental Theorem of Calculus Part 1: Integrals and Antiderivatives<\/h2>\r\n<p id=\"fs-id1170571639757\">As mentioned earlier, the Fundamental Theorem of Calculus is an extremely powerful theorem that establishes the relationship between differentiation and integration, and gives us a way to evaluate definite integrals without using Riemann sums or calculating areas. The theorem is comprised of two parts, the first of which, the Fundamental Theorem of Calculus, Part 1, is stated here. Part 1 establishes the relationship between differentiation and integration.<\/p>\r\n\r\n<div id=\"fs-id1170571704350\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Fundamental Theorem of Calculus, Part 1<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1170571679188\">If [latex]f(x)[\/latex] is continuous over an interval [latex]\\left[a,b\\right],[\/latex] and the function [latex]F(x)[\/latex] is defined by<\/p>\r\n\r\n<div id=\"fs-id1170572552076\" class=\"equation\" style=\"text-align: center;\">[latex]F(x)={\\displaystyle\\int }_{a}^{x}f(t)dt,[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572307182\">then [latex]{F}^{\\prime }(x)=f(x)[\/latex] over [latex]\\left[a,b\\right].[\/latex]<\/p>\r\n\r\n<\/div>\r\n<p id=\"fs-id1170572114562\">Before we delve into the proof, a couple of subtleties are worth mentioning here. First, a comment on the notation. Note that we have defined a function, [latex]F(x),[\/latex] as the definite integral of another function, [latex]f(t),[\/latex] from the point [latex]a[\/latex] to the point [latex]x[\/latex]. At first glance, this is confusing, because we have said several times that a definite integral is a number, and here it looks like it\u2019s a function. The key here is to notice that for any particular value of [latex]x[\/latex], the definite integral is a number. So the function [latex]F(x)[\/latex] returns a number (the value of the definite integral) for each value of [latex]x[\/latex].<\/p>\r\n<p id=\"fs-id1170572480292\">Second, it is worth commenting on some of the key implications of this theorem. There is a reason it is called the <em>Fundamental<\/em> Theorem of Calculus. Not only does it establish a relationship between integration and differentiation, but also it guarantees that any integrable function has an antiderivative. Specifically, it guarantees that any continuous function has an antiderivative.<\/p>\r\n\r\n<div id=\"fs-id1170572101687\" class=\"bc-section section\">\r\n<h3>Proof<\/h3>\r\nApplying the definition of the derivative, we have\r\n<div id=\"fs-id1170572337215\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ \\\\ {F}^{\\prime }(x)\\hfill &amp; =\\underset{h\\to 0}{\\text{lim}}\\dfrac{F(x+h)-F(x)}{h}\\hfill \\\\ \\\\ &amp; =\\underset{h\\to 0}{\\text{lim}}\\dfrac{1}{h}\\left[{\\displaystyle\\int }_{a}^{x+h}f(t)dt-{\\displaystyle\\int }_{a}^{x}f(t)dt\\right]\\hfill \\\\ &amp; =\\underset{h\\to 0}{\\text{lim}}\\dfrac{1}{h}\\left[{\\displaystyle\\int }_{a}^{x+h}f(t)dt+{\\displaystyle\\int }_{x}^{a}f(t)dt\\right]\\hfill \\\\ &amp; =\\underset{h\\to 0}{\\text{lim}}\\dfrac{1}{h}{\\displaystyle\\int }_{x}^{x+h}f(t)dt.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572608050\">Looking carefully at this last expression, we see [latex]\\dfrac{1}{h}{\\displaystyle\\int }_{x}^{x+h}f(t)dt[\/latex] is just the average value of the function [latex]f(x)[\/latex] over the interval [latex]\\left[x,x+h\\right].[\/latex] Therefore, by the mean value theorem for integrals, there is some number [latex]c[\/latex] in [latex]\\left[x,x+h\\right][\/latex] such that<\/p>\r\n\r\n<div id=\"fs-id1170571566095\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{1}{h}{\\displaystyle\\int }_{x}^{x+h}f(x)dx=f(c).[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572246217\">In addition, since [latex]c[\/latex] is between [latex]x[\/latex] and [latex]x+h[\/latex], [latex]c[\/latex] approaches [latex]x[\/latex] as [latex]h[\/latex] approaches zero. Also, since [latex]f(x)[\/latex] is continuous, we have [latex]\\underset{h\\to 0}{\\text{lim}}f(c)=\\underset{c\\to x}{\\text{lim}}f(c)=f(x).[\/latex] Putting all these pieces together, we have<\/p>\r\n\r\n<div id=\"fs-id1170572481212\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ {F}^{\\prime }(x)\\hfill &amp; =\\underset{h\\to 0}{\\text{lim}}\\frac{1}{h}{\\displaystyle\\int }_{x}^{x+h}f(x)dx\\hfill \\\\ &amp; =\\underset{h\\to 0}{\\text{lim}}f(c)\\hfill \\\\ &amp; =f(x),\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572505513\">and the proof is complete.<\/p>\r\n<p id=\"fs-id1170572558418\">[latex]_\\blacksquare[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1170572601345\" class=\"textbook exercises\">\r\n<h3>Example: Finding a Derivative with the Fundamental Theorem of Calculus<\/h3>\r\n<p id=\"fs-id1170572498741\">Use the first part of the Fundamental Theorem of Calculus to find the derivative of<\/p>\r\n\r\n<div id=\"fs-id1170572492494\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]g(x)={\\displaystyle\\int }_{1}^{x}\\dfrac{1}{{t}^{3}+1}dt.[\/latex]<\/div>\r\n[reveal-answer q=\"fs-id1170572346816\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572346816\"]\r\n<p id=\"fs-id1170572346816\">According to the Fundamental Theorem of Calculus, the derivative is given by<\/p>\r\n\r\n<div id=\"fs-id1170572346820\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{g}^{\\prime }(x)=\\frac{1}{{x}^{3}+1}.[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170572099777\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nUse the Fundamental Theorem of Calculus, Part 1 to find the derivative of [latex]g(r)={\\displaystyle\\int }_{0}^{r}\\sqrt{{x}^{2}+4}dx.[\/latex]\r\n<div id=\"fs-id1170572099780\" class=\"exercise\">[reveal-answer q=\"fs-id1170571586152\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571586152\"]\r\n<p id=\"fs-id1170571586152\">[latex]{g}^{\\prime }(r)=\\sqrt{{r}^{2}+4}[\/latex]<\/p>\r\n\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1170572099767\">Follow the procedures from the last example to solve the problem.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170572134770\" class=\"textbook exercises\">\r\n<h3>example: Using the Fundamental Theorem and the Chain Rule to Calculate Derivatives<\/h3>\r\nLet [latex]F(x)={\\displaystyle\\int }_{1}^{\\sqrt{x}} \\sin tdt.[\/latex] Find [latex]{F}^{\\prime }(x).[\/latex]\r\n<div id=\"fs-id1170572512075\" class=\"exercise\">[reveal-answer q=\"fs-id1170572103655\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572103655\"]\r\n<p id=\"fs-id1170572103655\">Letting [latex]u(x)=\\sqrt{x},[\/latex] we have [latex]F(x)={\\displaystyle\\int }_{1}^{u(x)} \\sin tdt.[\/latex] Thus, by the Fundamental Theorem of Calculus and the chain rule,<\/p>\r\n\r\n<div id=\"fs-id1170572452450\" class=\"equation unnumbered\">[latex]\\begin{array}{}\\\\ {F}^{\\prime }(x)\\hfill &amp; = \\sin (u(x))\\frac{du}{dx}\\hfill \\\\ &amp; = \\sin (u(x))\u00b7(\\frac{1}{2}{x}^{-1\\text{\/}2})\\hfill \\\\ &amp; =\\frac{ \\sin \\sqrt{x}}{2\\sqrt{x}}.\\hfill \\end{array}[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571710606\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nLet [latex]F(x)={\\displaystyle\\int }_{1}^{{x}^{3}} \\cos tdt.[\/latex] Find [latex]{F}^{\\prime }(x).[\/latex]\r\n\r\n[reveal-answer q=\"3776209\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"3776209\"]\r\n\r\nUse the chain rule to solve the problem.\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1170572557217\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572557217\"]\r\n\r\n[latex]{F}^{\\prime }(x)=3{x}^{2} \\cos {x}^{3}[\/latex]\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1170572228878\" class=\"textbook exercises\">\r\n<h3>Example: Using the Fundamental Theorem of Calculus with Two Variable Limits of Integration<\/h3>\r\nLet [latex]F(x)={\\displaystyle\\int }_{x}^{2x}{t}^{3}dt.[\/latex] Find [latex]{F}^{\\prime }(x).[\/latex]\r\n<div id=\"fs-id1170572228880\" class=\"exercise\">[reveal-answer q=\"fs-id1170572142330\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572142330\"]\r\n<p id=\"fs-id1170572142330\">We have [latex]F(x)={\\displaystyle\\int }_{x}^{2x}{t}^{3}dt.[\/latex] Both limits of integration are variable, so we need to split this into two integrals. We get<\/p>\r\n\r\n<div id=\"fs-id1170572141627\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ F(x)\\hfill &amp; ={\\displaystyle\\int }_{x}^{2x}{t}^{3}dt\\hfill \\\\ &amp; ={\\displaystyle\\int }_{x}^{0}{t}^{3}dt+{\\displaystyle\\int }_{0}^{2x}{t}^{3}dt\\hfill \\\\ &amp; =\\text{\u2212}{\\displaystyle\\int }_{0}^{x}{t}^{3}dt+{\\displaystyle\\int }_{0}^{2x}{t}^{3}dt.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1170571610445\">Differentiating the first term, we obtain<\/p>\r\n\r\n<div id=\"fs-id1170572245919\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}\\left[\\text{\u2212}{\\displaystyle\\int }_{0}^{x}{t}^{3}dt\\right]=\\text{\u2212}{x}^{3}.[\/latex]<\/div>\r\n<p id=\"fs-id1170572305916\">Differentiating the second term, we first let [latex]u(x)=2x.[\/latex] Then,<\/p>\r\n\r\n<div id=\"fs-id1170572168674\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\frac{d}{dx}\\left[{\\displaystyle\\int }_{0}^{2x}{t}^{3}dt\\right]\\hfill &amp; =\\frac{d}{dx}\\left[{\\displaystyle\\int }_{0}^{u(x)}{t}^{3}dt\\right]\\hfill \\\\ &amp; ={(u(x))}^{3}\\frac{du}{dx}\\hfill \\\\ &amp; ={(2x)}^{3}\u00b72\\hfill \\\\ &amp; =16{x}^{3}.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1170572207052\">Thus,<\/p>\r\n\r\n<div id=\"fs-id1170572336986\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ {F}^{\\prime }(x)\\hfill &amp; =\\frac{d}{dx}\\left[\\text{\u2212}{\\displaystyle\\int }_{0}^{x}{t}^{3}dt\\right]+\\frac{d}{dx}\\left[{\\displaystyle\\int }_{0}^{2x}{t}^{3}dt\\right]\\hfill \\\\ &amp; =\\text{\u2212}{x}^{3}+16{x}^{3}\\hfill \\\\ &amp; =15{x}^{3}.\\hfill \\end{array}[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Using the Fundamental Theorem of Calculus with Two Variable Limits of Integration.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/UdsTNaiWmbs?controls=0&amp;start=707&amp;end=831&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266834\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266834\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/5.3TheFundamentalTheoremOfCalculus707to831_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.3 The Fundamental Theorem of Calculus\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1170572177853\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nLet [latex]F(x)={\\displaystyle\\int }_{x}^{{x}^{2}} \\cos tdt.[\/latex] Find [latex]{F}^{\\prime }(x).[\/latex]\r\n<div id=\"fs-id1170572168834\" class=\"exercise\">[reveal-answer q=\"fs-id1170572508003\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572508003\"]\r\n<p id=\"fs-id1170572508003\">[latex]{F}^{\\prime }(x)=2x \\cos {x}^{2}- \\cos x[\/latex]<\/p>\r\n\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1170572168653\">Use the procedures from the previous example to solve the problem.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]20404[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]20453[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Fundamental Theorem of Calculus, Part 2: The Evaluation Theorem<\/h2>\r\n<p id=\"fs-id1170571697314\">The Fundamental Theorem of Calculus, Part 2, is perhaps the most important theorem in calculus. After tireless efforts by mathematicians for approximately 500 years, new techniques emerged that provided scientists with the necessary tools to explain many phenomena. Using calculus, astronomers could finally determine distances in space and map planetary orbits. Everyday financial problems such as calculating marginal costs or predicting total profit could now be handled with simplicity and accuracy. Engineers could calculate the bending strength of materials or the three-dimensional motion of objects. Our view of the world was forever changed with calculus.<\/p>\r\n<p id=\"fs-id1170571697316\">After finding approximate areas by adding the areas of [latex]n[\/latex] rectangles, the application of this theorem is straightforward by comparison. It almost seems too simple that the area of an entire curved region can be calculated by just evaluating an antiderivative at the first and last endpoints of an interval.<\/p>\r\n\r\n<div id=\"fs-id1170571660076\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">The Fundamental Theorem of Calculus, Part 2<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1170572448137\">If [latex]f[\/latex] is continuous over the interval [latex]\\left[a,b\\right][\/latex] and [latex]F(x)[\/latex] is any antiderivative of [latex]f(x),[\/latex] then<\/p>\r\n\r\n<div id=\"fs-id1170572230004\" class=\"equation\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{a}^{b}f(x)dx=F(b)-F(a)[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<p id=\"fs-id1170572416003\">We often see the notation [latex]{F(x)|}_{a}^{b}[\/latex] to denote the expression [latex]F(b)-F(a).[\/latex] We use this vertical bar and associated limits [latex]a[\/latex] and [latex]b[\/latex] to indicate that we should evaluate the function [latex]F(x)[\/latex] at the upper limit (in this case, [latex]b[\/latex]), and subtract the value of the function [latex]F(x)[\/latex] evaluated at the lower limit (in this case, [latex]a[\/latex]).<\/p>\r\n<p id=\"fs-id1170572622222\">The Fundamental Theorem of Calculus, Part 2 (also known as the evaluation theorem) states that if we can find an antiderivative for the integrand, then we can evaluate the definite integral by evaluating the antiderivative at the endpoints of the interval and subtracting.<\/p>\r\n\r\n<div id=\"fs-id1170572430375\" class=\"bc-section section\">\r\n<h3>Proof<\/h3>\r\n<p id=\"fs-id1170572430381\">Let [latex]P=\\left\\{{x}_{i}\\right\\},i=0,1\\text{,\u2026,}n[\/latex] be a regular partition of [latex]\\left[a,b\\right].[\/latex] Then, we can write<\/p>\r\n\r\n<div id=\"fs-id1170572435583\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}F(b)-F(a)\\hfill &amp; =F({x}_{n})-F({x}_{0})\\hfill \\\\ &amp; =\\left[F({x}_{n})-F({x}_{n-1})\\right]+\\left[F({x}_{n-1})-F({x}_{n-2})\\right]+\\text{\u2026}+\\left[F({x}_{1})-F({x}_{0})\\right]\\hfill \\\\ \\\\ &amp; =\\underset{i=1}{\\overset{n}{\\text{\u2211}}}\\left[F({x}_{i})-F({x}_{i-1})\\right].\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572370648\">Now, we know <em>F<\/em> is an antiderivative of [latex]f[\/latex] over [latex]\\left[a,b\\right],[\/latex] so by the Mean Value Theorem (see <a class=\"target-chapter\" href=\"https:\/\/courses.lumenlearning.com\/suny-openstax-calculus1\/chapter\/the-mean-value-theorem\/\">The Mean Value Theorem<\/a>) for [latex]i=0,1\\text{,\u2026,}n[\/latex] we can find [latex]{c}_{i}[\/latex] in [latex]\\left[{x}_{i-1},{x}_{i}\\right][\/latex] such that<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]F({x}_{i})-F({x}_{i-1})={F}^{\\prime }({c}_{i})({x}_{i}-{x}_{i-1})=f({c}_{i})\\text{\u0394}x.[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170571699032\">Then, substituting into the previous equation, we have<\/p>\r\n\r\n<div id=\"fs-id1170571699036\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]F(b)-F(a)=\\underset{i=1}{\\overset{n}{\\text{\u2211}}}f({c}_{i})\\text{\u0394}x.[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572554028\">Taking the limit of both sides as [latex]n\\to \\infty ,[\/latex] we obtain<\/p>\r\n\r\n<div id=\"fs-id1170572548857\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ F(b)-F(a)\\hfill &amp; =\\underset{n\\to \\infty }{\\text{lim}}\\underset{i=1}{\\overset{n}{\\text{\u2211}}}f({c}_{i})\\text{\u0394}x\\hfill \\\\ &amp; ={\\displaystyle\\int }_{a}^{b}f(x)dx.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1170571678907\">[latex]_\\blacksquare[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1170571678911\" class=\"textbook exercises\">\r\n<h3>Example: Evaluating an Integral with the Fundamental Theorem of Calculus<\/h3>\r\n<p id=\"fs-id1170571678920\">Use the second part of the Fundamental Theorem of Calculus to evaluate<\/p>\r\n\r\n<div id=\"fs-id1170572330418\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{-2}^{2}({t}^{2}-4)dt.[\/latex]<\/div>\r\n<div id=\"fs-id1170571678913\" class=\"exercise\">\r\n\r\n[reveal-answer q=\"fs-id1170572173704\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572173704\"]\r\n<p id=\"fs-id1170572173704\">Recall the power rule for <a class=\"target-chapter\" href=\"https:\/\/courses.lumenlearning.com\/suny-openstax-calculus1\/chapter\/antiderivatives\/\">Antiderivatives<\/a>:<\/p>\r\n\r\n<div id=\"fs-id1170572547894\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{ If }y={x}^{n},\\displaystyle\\int {x}^{n}dx=\\frac{{x}^{n+1}}{n+1}+C.[\/latex]<\/div>\r\n<p id=\"fs-id1170571697070\">Use this rule to find the antiderivative of the function and then apply the theorem. We have<\/p>\r\n\r\n<div id=\"fs-id1170571719649\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int }_{-2}^{2}({t}^{2}-4)dt\\hfill &amp; =\\frac{{t}^{3}}{3}-{4t|}_{-2}^{2}\\hfill \\\\ \\\\ \\\\ &amp; =\\left[\\frac{{(2)}^{3}}{3}-4(2)\\right]-\\left[\\frac{{(-2)}^{3}}{3}-4(-2)\\right]\\hfill \\\\ &amp; =(\\frac{8}{3}-8)-(-\\frac{8}{3}+8)\\hfill \\\\ &amp; =\\frac{8}{3}-8+\\frac{8}{3}-8\\hfill \\\\ &amp; =\\frac{16}{3}-16\\hfill \\\\ &amp; =-\\frac{32}{3}.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1170572135831\"><strong>Analysis<\/strong><\/p>\r\n<p id=\"fs-id1170572135837\">Notice that we did not include the \u201c+ <em>C<\/em>\u201d term when we wrote the antiderivative. The reason is that, according to the Fundamental Theorem of Calculus, Part 2, <em>any<\/em> antiderivative works. So, for convenience, we chose the antiderivative with [latex]C=0.[\/latex] If we had chosen another antiderivative, the constant term would have canceled out. This always happens when evaluating a definite integral.<\/p>\r\n<p id=\"fs-id1170571660070\">The region of the area we just calculated is depicted in Figure 3. Note that the region between the curve and the [latex]x[\/latex]-axis is all below the [latex]x[\/latex]-axis. Area is always positive, but a definite integral can still produce a negative number (a net signed area). For example, if this were a profit function, a negative number indicates the company is operating at a loss over the given interval.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11204110\/CNX_Calc_Figure_05_03_004.jpg\" alt=\"The graph of the parabola f(t) = t^2 \u2013 4 over [-4, 4]. The area above the curve and below the x axis over [-2, 2] is shaded.\" width=\"325\" height=\"283\" \/> Figure 3. The evaluation of a definite integral can produce a negative value, even though area is always positive.[\/caption][\/hidden-answer]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571561288\" class=\"textbook exercises\">\r\n<h3>Example: Evaluating a Definite Integral Using the Fundamental Theorem of Calculus, Part 2<\/h3>\r\n<p id=\"fs-id1170572607951\">Evaluate the following integral using the Fundamental Theorem of Calculus, Part 2:<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{1}^{9}\\dfrac{x-1}{\\sqrt{x}}dx.[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572130389\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572130389\"]\r\n<p id=\"fs-id1170572130389\">First, eliminate the radical by rewriting the integral using rational exponents. Then, separate the numerator terms by writing each one over the denominator:<\/p>\r\n\r\n<div id=\"fs-id1170572130394\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{1}^{9}\\frac{x-1}{{x}^{1\\text{\/}2}}dx={\\displaystyle\\int }_{1}^{9}(\\frac{x}{{x}^{1\\text{\/}2}}-\\frac{1}{{x}^{1\\text{\/}2}})dx\\text{.}[\/latex]<\/div>\r\n<p id=\"fs-id1170572624682\">Use the properties of exponents to simplify:<\/p>\r\n\r\n<div id=\"fs-id1170572233529\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{1}^{9}(\\frac{x}{{x}^{1\\text{\/}2}}-\\frac{1}{{x}^{1\\text{\/}2}})dx={\\displaystyle\\int }_{1}^{9}({x}^{1\\text{\/}2}-{x}^{-1\\text{\/}2})dx\\text{.}[\/latex]<\/div>\r\n<p id=\"fs-id1170572563042\">Now, integrate using the power rule:<\/p>\r\n\r\n<div id=\"fs-id1170572563045\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ {\\displaystyle\\int }_{1}^{9}({x}^{1\\text{\/}2}-{x}^{-1\\text{\/}2})dx\\hfill &amp; ={(\\frac{{x}^{3\\text{\/}2}}{\\frac{3}{2}}-\\frac{{x}^{1\\text{\/}2}}{\\frac{1}{2}})|}_{1}^{9}\\hfill \\\\ \\\\ &amp; =\\left[\\frac{{(9)}^{3\\text{\/}2}}{\\frac{3}{2}}-\\frac{{(9)}^{1\\text{\/}2}}{\\frac{1}{2}}\\right]-\\left[\\frac{{(1)}^{3\\text{\/}2}}{\\frac{3}{2}}-\\frac{{(1)}^{1\\text{\/}2}}{\\frac{1}{2}}\\right]\\hfill \\\\ &amp; =\\left[\\frac{2}{3}(27)-2(3)\\right]-\\left[\\frac{2}{3}(1)-2(1)\\right]\\hfill \\\\ &amp; =18-6-\\frac{2}{3}+2\\hfill \\\\ &amp; =\\frac{40}{3}.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1170572522399\">See Figure 4.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11204112\/CNX_Calc_Figure_05_03_005.jpg\" alt=\"The graph of the function f(x) = (x-1) \/ sqrt(x) over [0,9]. The area under the graph over [1,9] is shaded.\" width=\"325\" height=\"246\" \/> Figure 4. The area under the curve from [latex]x=1[\/latex] to [latex]x=9[\/latex] can be calculated by evaluating a definite integral.[\/caption]\r\n<div class=\"wp-caption-text\"><\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Evaluating a Definite Integral Using the Fundamental Theorem of Calculus, Part 2.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/UdsTNaiWmbs?controls=0&amp;start=998&amp;end=1063&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266833\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266833\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/5.3TheFundamentalTheoremOfCalculus998to1063_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.3 The Fundamental Theorem of Calculus\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1170571609442\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nUse the second part of the Fundamental Theorem of Calculus to evaluate [latex]{\\displaystyle\\int }_{1}^{2}{x}^{-4}dx.[\/latex]\r\n\r\n[reveal-answer q=\"3887125\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"3887125\"]\r\n\r\nUse the power rule.\r\n\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1170571639103\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571639103\"]\r\n\r\n[latex]\\frac{7}{24}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]211336[\/ohm_question]\r\n\r\n<\/div>\r\n<div id=\"fs-id1170571597325\" class=\"textbook exercises\">\r\n<h3>Example: A Roller-Skating Race<\/h3>\r\nJames and Kathy are racing on roller skates. They race along a long, straight track, and whoever has gone the farthest after 5 sec wins a prize. If James can skate at a velocity of [latex]f(t)=5+2t[\/latex] ft\/sec and Kathy can skate at a velocity of [latex]g(t)=10+ \\cos (\\frac{\\pi }{2}t)[\/latex] ft\/sec, who is going to win the race?\r\n<div id=\"fs-id1170571597327\" class=\"exercise\">[reveal-answer q=\"fs-id1170572332613\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572332613\"]\r\n<p id=\"fs-id1170572332613\">We need to integrate both functions over the interval [latex]\\left[0,5\\right][\/latex] and see which value is bigger. For James, we want to calculate<\/p>\r\n\r\n<div id=\"fs-id1170572332632\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{0}^{5}(5+2t)dt.[\/latex]<\/div>\r\n<p id=\"fs-id1170571758991\">Using the power rule, we have<\/p>\r\n\r\n<div id=\"fs-id1170571758994\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int }_{0}^{5}(5+2t)dt\\hfill &amp; ={(5t+{t}^{2})|}_{0}^{5}\\hfill \\\\ &amp; =(25+25)=50.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1170571772586\">Thus, James has skated 50 ft after 5 sec. Turning now to Kathy, we want to calculate<\/p>\r\n\r\n<div id=\"fs-id1170571772589\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{0}^{5}10+ \\cos (\\frac{\\pi }{2}t)dt.[\/latex]<\/div>\r\n<p id=\"fs-id1170572292292\">We know [latex] \\sin t[\/latex] is an antiderivative of [latex] \\cos t,[\/latex] so it is reasonable to expect that an antiderivative of [latex] \\cos (\\frac{\\pi }{2}t)[\/latex] would involve [latex] \\sin (\\frac{\\pi }{2}t).[\/latex] However, when we differentiate [latex] \\sin (\\frac{\\pi }{2}t),[\/latex] we get [latex]\\frac{\\pi }{2} \\cos (\\frac{\\pi }{2}t)[\/latex] as a result of the chain rule, so we have to account for this additional coefficient when we integrate. We obtain<\/p>\r\n\r\n<div id=\"fs-id1170571599308\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int }_{0}^{5}10+ \\cos (\\frac{\\pi }{2}t)dt\\hfill &amp; ={(10t+\\frac{2}{\\pi } \\sin (\\frac{\\pi }{2}t))|}_{0}^{5}\\hfill \\\\ &amp; =(50+\\frac{2}{\\pi })-(0-\\frac{2}{\\pi } \\sin 0)\\hfill \\\\ &amp; \\approx 50.6.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1170572368405\">Kathy has skated approximately 50.6 ft after 5 sec. Kathy wins, but not by much!<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: A Roller-Skating Race.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/UdsTNaiWmbs?controls=0&amp;start=1072&amp;end=1258&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266835\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266835\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/5.3TheFundamentalTheoremOfCalculus1072to1258_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.3 The Fundamental Theorem of Calculus\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1170572368412\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1170572368419\">Suppose James and Kathy have a rematch, but this time the official stops the contest after only 3 sec. Does this change the outcome?<\/p>\r\n[reveal-answer q=\"fs-id1170571773427\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571773427\"]\r\n<p id=\"fs-id1170571773427\">Kathy still wins, but by a much larger margin: James skates 24 ft in 3 sec, but Kathy skates 29.3634 ft in 3 sec.<\/p>\r\n\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1170572368423\">Change the limits of integration from those in the previous example.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1170571773434\" class=\"textbox tryit\">\r\n<h3>Activity: A Parachutist in Free Fall<\/h3>\r\n&nbsp;\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"701\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11204117\/CNX_Calc_Figure_05_03_006.jpg\" alt=\"Two skydivers free falling in the sky.\" width=\"701\" height=\"462\" \/> Figure 5. Skydivers can adjust the velocity of their dive by changing the position of their body during the free fall. (credit: Jeremy T. Lock)[\/caption]\r\n\r\n&nbsp;\r\n<p id=\"fs-id1170572218426\">Julie is an avid <span class=\"no-emphasis\">skydiver<\/span>. She has more than 300 jumps under her belt and has mastered the art of making adjustments to her body position in the air to control how fast she falls. If she arches her back and points her belly toward the ground, she reaches a terminal velocity of approximately 120 mph (176 ft\/sec). If, instead, she orients her body with her head straight down, she falls faster, reaching a terminal velocity of 150 mph (220 ft\/sec).<\/p>\r\n<p id=\"fs-id1170572218437\">Since Julie will be moving (falling) in a downward direction, we assume the downward direction is positive to simplify our calculations. Julie executes her jumps from an altitude of 12,500 ft. After she exits the aircraft, she immediately starts falling at a velocity given by [latex]v(t)=32t.[\/latex] She continues to accelerate according to this velocity function until she reaches terminal velocity. After she reaches terminal velocity, her speed remains constant until she pulls her ripcord and slows down to land.<\/p>\r\n<p id=\"fs-id1170572419103\">On her first jump of the day, Julie orients herself in the slower \u201cbelly down\u201d position (terminal velocity is 176 ft\/sec). Using this information, answer the following questions.<\/p>\r\n\r\n<ol id=\"fs-id1170572419110\">\r\n \t<li>How long after she exits the aircraft does Julie reach terminal velocity?<\/li>\r\n \t<li>Based on your answer to question 1, set up an expression involving one or more integrals that represents the distance Julie falls after 30 sec.<\/li>\r\n \t<li>If Julie pulls her ripcord at an altitude of 3000 ft, how long does she spend in a free fall?<\/li>\r\n \t<li>Julie pulls her ripcord at 3000 ft. It takes 5 sec for her parachute to open completely and for her to slow down, during which time she falls another 400 ft. After her canopy is fully open, her speed is reduced to 16 ft\/sec. Find the total time Julie spends in the air, from the time she leaves the airplane until the time her feet touch the ground.\r\nOn Julie\u2019s second jump of the day, she decides she wants to fall a little faster and orients herself in the \u201chead down\u201d position. Her terminal velocity in this position is 220 ft\/sec. Answer these questions based on this velocity:<\/li>\r\n \t<li>How long does it take Julie to reach terminal velocity in this case?<\/li>\r\n \t<li>Before pulling her ripcord, Julie reorients her body in the \u201cbelly down\u201d position so she is not moving quite as fast when her parachute opens. If she begins this maneuver at an altitude of 4000 ft, how long does she spend in a free fall before beginning the reorientation?\r\nSome jumpers wear \u201c<span class=\"no-emphasis\">wingsuits<\/span>\u201d (see Figure 6). These suits have fabric panels between the arms and legs and allow the wearer to glide around in a free fall, much like a flying squirrel. (Indeed, the suits are sometimes called \u201cflying squirrel suits.\u201d) When wearing these suits, terminal velocity can be reduced to about 30 mph (44 ft\/sec), allowing the wearers a much longer time in the air. Wingsuit flyers still use parachutes to land; although the vertical velocities are within the margin of safety, horizontal velocities can exceed 70 mph, much too fast to land safely.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"700\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11204124\/CNX_Calc_Figure_05_03_007.jpg\" alt=\"A person falling in a wingsuit, which works to reduce the vertical velocity of a skydiver\u2019s fall.\" width=\"700\" height=\"392\" \/> Figure 6. The fabric panels on the arms and legs of a wingsuit work to reduce the vertical velocity of a skydiver\u2019s fall. (credit: Richard Schneider)[\/caption]<\/li>\r\n<\/ol>\r\n<p id=\"fs-id1163722683509\">Answer the following question based on the velocity in a wingsuit.<\/p>\r\n\r\n<ol id=\"fs-id1163722683512\" start=\"7\">\r\n \t<li>If Julie dons a wingsuit before her third jump of the day, and she pulls her ripcord at an altitude of 3000 ft, how long does she get to spend gliding around in the air?<\/li>\r\n<\/ol>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>State the meaning of the Fundamental Theorem of Calculus, Part 1<\/li>\n<li>Use the Fundamental Theorem of Calculus, Part 1, to evaluate derivatives of integrals<\/li>\n<li>State the meaning of the Fundamental Theorem of Calculus, Part 2<\/li>\n<li>Use the Fundamental Theorem of Calculus, Part 2, to evaluate definite integrals<\/li>\n<li>Explain the relationship between differentiation and integration<\/li>\n<\/ul>\n<\/div>\n<h2>Fundamental Theorem of Calculus Part 1: Integrals and Antiderivatives<\/h2>\n<p id=\"fs-id1170571639757\">As mentioned earlier, the Fundamental Theorem of Calculus is an extremely powerful theorem that establishes the relationship between differentiation and integration, and gives us a way to evaluate definite integrals without using Riemann sums or calculating areas. The theorem is comprised of two parts, the first of which, the Fundamental Theorem of Calculus, Part 1, is stated here. Part 1 establishes the relationship between differentiation and integration.<\/p>\n<div id=\"fs-id1170571704350\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Fundamental Theorem of Calculus, Part 1<\/h3>\n<hr \/>\n<p id=\"fs-id1170571679188\">If [latex]f(x)[\/latex] is continuous over an interval [latex]\\left[a,b\\right],[\/latex] and the function [latex]F(x)[\/latex] is defined by<\/p>\n<div id=\"fs-id1170572552076\" class=\"equation\" style=\"text-align: center;\">[latex]F(x)={\\displaystyle\\int }_{a}^{x}f(t)dt,[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572307182\">then [latex]{F}^{\\prime }(x)=f(x)[\/latex] over [latex]\\left[a,b\\right].[\/latex]<\/p>\n<\/div>\n<p id=\"fs-id1170572114562\">Before we delve into the proof, a couple of subtleties are worth mentioning here. First, a comment on the notation. Note that we have defined a function, [latex]F(x),[\/latex] as the definite integral of another function, [latex]f(t),[\/latex] from the point [latex]a[\/latex] to the point [latex]x[\/latex]. At first glance, this is confusing, because we have said several times that a definite integral is a number, and here it looks like it\u2019s a function. The key here is to notice that for any particular value of [latex]x[\/latex], the definite integral is a number. So the function [latex]F(x)[\/latex] returns a number (the value of the definite integral) for each value of [latex]x[\/latex].<\/p>\n<p id=\"fs-id1170572480292\">Second, it is worth commenting on some of the key implications of this theorem. There is a reason it is called the <em>Fundamental<\/em> Theorem of Calculus. Not only does it establish a relationship between integration and differentiation, but also it guarantees that any integrable function has an antiderivative. Specifically, it guarantees that any continuous function has an antiderivative.<\/p>\n<div id=\"fs-id1170572101687\" class=\"bc-section section\">\n<h3>Proof<\/h3>\n<p>Applying the definition of the derivative, we have<\/p>\n<div id=\"fs-id1170572337215\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ \\\\ {F}^{\\prime }(x)\\hfill & =\\underset{h\\to 0}{\\text{lim}}\\dfrac{F(x+h)-F(x)}{h}\\hfill \\\\ \\\\ & =\\underset{h\\to 0}{\\text{lim}}\\dfrac{1}{h}\\left[{\\displaystyle\\int }_{a}^{x+h}f(t)dt-{\\displaystyle\\int }_{a}^{x}f(t)dt\\right]\\hfill \\\\ & =\\underset{h\\to 0}{\\text{lim}}\\dfrac{1}{h}\\left[{\\displaystyle\\int }_{a}^{x+h}f(t)dt+{\\displaystyle\\int }_{x}^{a}f(t)dt\\right]\\hfill \\\\ & =\\underset{h\\to 0}{\\text{lim}}\\dfrac{1}{h}{\\displaystyle\\int }_{x}^{x+h}f(t)dt.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572608050\">Looking carefully at this last expression, we see [latex]\\dfrac{1}{h}{\\displaystyle\\int }_{x}^{x+h}f(t)dt[\/latex] is just the average value of the function [latex]f(x)[\/latex] over the interval [latex]\\left[x,x+h\\right].[\/latex] Therefore, by the mean value theorem for integrals, there is some number [latex]c[\/latex] in [latex]\\left[x,x+h\\right][\/latex] such that<\/p>\n<div id=\"fs-id1170571566095\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{1}{h}{\\displaystyle\\int }_{x}^{x+h}f(x)dx=f(c).[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572246217\">In addition, since [latex]c[\/latex] is between [latex]x[\/latex] and [latex]x+h[\/latex], [latex]c[\/latex] approaches [latex]x[\/latex] as [latex]h[\/latex] approaches zero. Also, since [latex]f(x)[\/latex] is continuous, we have [latex]\\underset{h\\to 0}{\\text{lim}}f(c)=\\underset{c\\to x}{\\text{lim}}f(c)=f(x).[\/latex] Putting all these pieces together, we have<\/p>\n<div id=\"fs-id1170572481212\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ {F}^{\\prime }(x)\\hfill & =\\underset{h\\to 0}{\\text{lim}}\\frac{1}{h}{\\displaystyle\\int }_{x}^{x+h}f(x)dx\\hfill \\\\ & =\\underset{h\\to 0}{\\text{lim}}f(c)\\hfill \\\\ & =f(x),\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572505513\">and the proof is complete.<\/p>\n<p id=\"fs-id1170572558418\">[latex]_\\blacksquare[\/latex]<\/p>\n<div id=\"fs-id1170572601345\" class=\"textbook exercises\">\n<h3>Example: Finding a Derivative with the Fundamental Theorem of Calculus<\/h3>\n<p id=\"fs-id1170572498741\">Use the first part of the Fundamental Theorem of Calculus to find the derivative of<\/p>\n<div id=\"fs-id1170572492494\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]g(x)={\\displaystyle\\int }_{1}^{x}\\dfrac{1}{{t}^{3}+1}dt.[\/latex]<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572346816\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572346816\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572346816\">According to the Fundamental Theorem of Calculus, the derivative is given by<\/p>\n<div id=\"fs-id1170572346820\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{g}^{\\prime }(x)=\\frac{1}{{x}^{3}+1}.[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170572099777\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Use the Fundamental Theorem of Calculus, Part 1 to find the derivative of [latex]g(r)={\\displaystyle\\int }_{0}^{r}\\sqrt{{x}^{2}+4}dx.[\/latex]<\/p>\n<div id=\"fs-id1170572099780\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571586152\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571586152\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571586152\">[latex]{g}^{\\prime }(r)=\\sqrt{{r}^{2}+4}[\/latex]<\/p>\n<h4>Hint<\/h4>\n<p id=\"fs-id1170572099767\">Follow the procedures from the last example to solve the problem.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170572134770\" class=\"textbook exercises\">\n<h3>example: Using the Fundamental Theorem and the Chain Rule to Calculate Derivatives<\/h3>\n<p>Let [latex]F(x)={\\displaystyle\\int }_{1}^{\\sqrt{x}} \\sin tdt.[\/latex] Find [latex]{F}^{\\prime }(x).[\/latex]<\/p>\n<div id=\"fs-id1170572512075\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572103655\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572103655\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572103655\">Letting [latex]u(x)=\\sqrt{x},[\/latex] we have [latex]F(x)={\\displaystyle\\int }_{1}^{u(x)} \\sin tdt.[\/latex] Thus, by the Fundamental Theorem of Calculus and the chain rule,<\/p>\n<div id=\"fs-id1170572452450\" class=\"equation unnumbered\">[latex]\\begin{array}{}\\\\ {F}^{\\prime }(x)\\hfill & = \\sin (u(x))\\frac{du}{dx}\\hfill \\\\ & = \\sin (u(x))\u00b7(\\frac{1}{2}{x}^{-1\\text{\/}2})\\hfill \\\\ & =\\frac{ \\sin \\sqrt{x}}{2\\sqrt{x}}.\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571710606\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Let [latex]F(x)={\\displaystyle\\int }_{1}^{{x}^{3}} \\cos tdt.[\/latex] Find [latex]{F}^{\\prime }(x).[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q3776209\">Hint<\/span><\/p>\n<div id=\"q3776209\" class=\"hidden-answer\" style=\"display: none\">\n<p>Use the chain rule to solve the problem.\n<\/p><\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572557217\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572557217\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]{F}^{\\prime }(x)=3{x}^{2} \\cos {x}^{3}[\/latex]\n<\/p><\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170572228878\" class=\"textbook exercises\">\n<h3>Example: Using the Fundamental Theorem of Calculus with Two Variable Limits of Integration<\/h3>\n<p>Let [latex]F(x)={\\displaystyle\\int }_{x}^{2x}{t}^{3}dt.[\/latex] Find [latex]{F}^{\\prime }(x).[\/latex]<\/p>\n<div id=\"fs-id1170572228880\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572142330\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572142330\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572142330\">We have [latex]F(x)={\\displaystyle\\int }_{x}^{2x}{t}^{3}dt.[\/latex] Both limits of integration are variable, so we need to split this into two integrals. We get<\/p>\n<div id=\"fs-id1170572141627\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ F(x)\\hfill & ={\\displaystyle\\int }_{x}^{2x}{t}^{3}dt\\hfill \\\\ & ={\\displaystyle\\int }_{x}^{0}{t}^{3}dt+{\\displaystyle\\int }_{0}^{2x}{t}^{3}dt\\hfill \\\\ & =\\text{\u2212}{\\displaystyle\\int }_{0}^{x}{t}^{3}dt+{\\displaystyle\\int }_{0}^{2x}{t}^{3}dt.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170571610445\">Differentiating the first term, we obtain<\/p>\n<div id=\"fs-id1170572245919\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}\\left[\\text{\u2212}{\\displaystyle\\int }_{0}^{x}{t}^{3}dt\\right]=\\text{\u2212}{x}^{3}.[\/latex]<\/div>\n<p id=\"fs-id1170572305916\">Differentiating the second term, we first let [latex]u(x)=2x.[\/latex] Then,<\/p>\n<div id=\"fs-id1170572168674\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\frac{d}{dx}\\left[{\\displaystyle\\int }_{0}^{2x}{t}^{3}dt\\right]\\hfill & =\\frac{d}{dx}\\left[{\\displaystyle\\int }_{0}^{u(x)}{t}^{3}dt\\right]\\hfill \\\\ & ={(u(x))}^{3}\\frac{du}{dx}\\hfill \\\\ & ={(2x)}^{3}\u00b72\\hfill \\\\ & =16{x}^{3}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170572207052\">Thus,<\/p>\n<div id=\"fs-id1170572336986\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ {F}^{\\prime }(x)\\hfill & =\\frac{d}{dx}\\left[\\text{\u2212}{\\displaystyle\\int }_{0}^{x}{t}^{3}dt\\right]+\\frac{d}{dx}\\left[{\\displaystyle\\int }_{0}^{2x}{t}^{3}dt\\right]\\hfill \\\\ & =\\text{\u2212}{x}^{3}+16{x}^{3}\\hfill \\\\ & =15{x}^{3}.\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Using the Fundamental Theorem of Calculus with Two Variable Limits of Integration.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/UdsTNaiWmbs?controls=0&amp;start=707&amp;end=831&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266834\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266834\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/5.3TheFundamentalTheoremOfCalculus707to831_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.3 The Fundamental Theorem of Calculus&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1170572177853\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Let [latex]F(x)={\\displaystyle\\int }_{x}^{{x}^{2}} \\cos tdt.[\/latex] Find [latex]{F}^{\\prime }(x).[\/latex]<\/p>\n<div id=\"fs-id1170572168834\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572508003\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572508003\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572508003\">[latex]{F}^{\\prime }(x)=2x \\cos {x}^{2}- \\cos x[\/latex]<\/p>\n<h4>Hint<\/h4>\n<p id=\"fs-id1170572168653\">Use the procedures from the previous example to solve the problem.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm20404\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=20404&theme=oea&iframe_resize_id=ohm20404&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm20453\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=20453&theme=oea&iframe_resize_id=ohm20453&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Fundamental Theorem of Calculus, Part 2: The Evaluation Theorem<\/h2>\n<p id=\"fs-id1170571697314\">The Fundamental Theorem of Calculus, Part 2, is perhaps the most important theorem in calculus. After tireless efforts by mathematicians for approximately 500 years, new techniques emerged that provided scientists with the necessary tools to explain many phenomena. Using calculus, astronomers could finally determine distances in space and map planetary orbits. Everyday financial problems such as calculating marginal costs or predicting total profit could now be handled with simplicity and accuracy. Engineers could calculate the bending strength of materials or the three-dimensional motion of objects. Our view of the world was forever changed with calculus.<\/p>\n<p id=\"fs-id1170571697316\">After finding approximate areas by adding the areas of [latex]n[\/latex] rectangles, the application of this theorem is straightforward by comparison. It almost seems too simple that the area of an entire curved region can be calculated by just evaluating an antiderivative at the first and last endpoints of an interval.<\/p>\n<div id=\"fs-id1170571660076\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">The Fundamental Theorem of Calculus, Part 2<\/h3>\n<hr \/>\n<p id=\"fs-id1170572448137\">If [latex]f[\/latex] is continuous over the interval [latex]\\left[a,b\\right][\/latex] and [latex]F(x)[\/latex] is any antiderivative of [latex]f(x),[\/latex] then<\/p>\n<div id=\"fs-id1170572230004\" class=\"equation\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{a}^{b}f(x)dx=F(b)-F(a)[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<p id=\"fs-id1170572416003\">We often see the notation [latex]{F(x)|}_{a}^{b}[\/latex] to denote the expression [latex]F(b)-F(a).[\/latex] We use this vertical bar and associated limits [latex]a[\/latex] and [latex]b[\/latex] to indicate that we should evaluate the function [latex]F(x)[\/latex] at the upper limit (in this case, [latex]b[\/latex]), and subtract the value of the function [latex]F(x)[\/latex] evaluated at the lower limit (in this case, [latex]a[\/latex]).<\/p>\n<p id=\"fs-id1170572622222\">The Fundamental Theorem of Calculus, Part 2 (also known as the evaluation theorem) states that if we can find an antiderivative for the integrand, then we can evaluate the definite integral by evaluating the antiderivative at the endpoints of the interval and subtracting.<\/p>\n<div id=\"fs-id1170572430375\" class=\"bc-section section\">\n<h3>Proof<\/h3>\n<p id=\"fs-id1170572430381\">Let [latex]P=\\left\\{{x}_{i}\\right\\},i=0,1\\text{,\u2026,}n[\/latex] be a regular partition of [latex]\\left[a,b\\right].[\/latex] Then, we can write<\/p>\n<div id=\"fs-id1170572435583\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}F(b)-F(a)\\hfill & =F({x}_{n})-F({x}_{0})\\hfill \\\\ & =\\left[F({x}_{n})-F({x}_{n-1})\\right]+\\left[F({x}_{n-1})-F({x}_{n-2})\\right]+\\text{\u2026}+\\left[F({x}_{1})-F({x}_{0})\\right]\\hfill \\\\ \\\\ & =\\underset{i=1}{\\overset{n}{\\text{\u2211}}}\\left[F({x}_{i})-F({x}_{i-1})\\right].\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572370648\">Now, we know <em>F<\/em> is an antiderivative of [latex]f[\/latex] over [latex]\\left[a,b\\right],[\/latex] so by the Mean Value Theorem (see <a class=\"target-chapter\" href=\"https:\/\/courses.lumenlearning.com\/suny-openstax-calculus1\/chapter\/the-mean-value-theorem\/\">The Mean Value Theorem<\/a>) for [latex]i=0,1\\text{,\u2026,}n[\/latex] we can find [latex]{c}_{i}[\/latex] in [latex]\\left[{x}_{i-1},{x}_{i}\\right][\/latex] such that<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]F({x}_{i})-F({x}_{i-1})={F}^{\\prime }({c}_{i})({x}_{i}-{x}_{i-1})=f({c}_{i})\\text{\u0394}x.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571699032\">Then, substituting into the previous equation, we have<\/p>\n<div id=\"fs-id1170571699036\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]F(b)-F(a)=\\underset{i=1}{\\overset{n}{\\text{\u2211}}}f({c}_{i})\\text{\u0394}x.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572554028\">Taking the limit of both sides as [latex]n\\to \\infty ,[\/latex] we obtain<\/p>\n<div id=\"fs-id1170572548857\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ F(b)-F(a)\\hfill & =\\underset{n\\to \\infty }{\\text{lim}}\\underset{i=1}{\\overset{n}{\\text{\u2211}}}f({c}_{i})\\text{\u0394}x\\hfill \\\\ & ={\\displaystyle\\int }_{a}^{b}f(x)dx.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170571678907\">[latex]_\\blacksquare[\/latex]<\/p>\n<div id=\"fs-id1170571678911\" class=\"textbook exercises\">\n<h3>Example: Evaluating an Integral with the Fundamental Theorem of Calculus<\/h3>\n<p id=\"fs-id1170571678920\">Use the second part of the Fundamental Theorem of Calculus to evaluate<\/p>\n<div id=\"fs-id1170572330418\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{-2}^{2}({t}^{2}-4)dt.[\/latex]<\/div>\n<div id=\"fs-id1170571678913\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572173704\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572173704\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572173704\">Recall the power rule for <a class=\"target-chapter\" href=\"https:\/\/courses.lumenlearning.com\/suny-openstax-calculus1\/chapter\/antiderivatives\/\">Antiderivatives<\/a>:<\/p>\n<div id=\"fs-id1170572547894\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{ If }y={x}^{n},\\displaystyle\\int {x}^{n}dx=\\frac{{x}^{n+1}}{n+1}+C.[\/latex]<\/div>\n<p id=\"fs-id1170571697070\">Use this rule to find the antiderivative of the function and then apply the theorem. We have<\/p>\n<div id=\"fs-id1170571719649\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int }_{-2}^{2}({t}^{2}-4)dt\\hfill & =\\frac{{t}^{3}}{3}-{4t|}_{-2}^{2}\\hfill \\\\ \\\\ \\\\ & =\\left[\\frac{{(2)}^{3}}{3}-4(2)\\right]-\\left[\\frac{{(-2)}^{3}}{3}-4(-2)\\right]\\hfill \\\\ & =(\\frac{8}{3}-8)-(-\\frac{8}{3}+8)\\hfill \\\\ & =\\frac{8}{3}-8+\\frac{8}{3}-8\\hfill \\\\ & =\\frac{16}{3}-16\\hfill \\\\ & =-\\frac{32}{3}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170572135831\"><strong>Analysis<\/strong><\/p>\n<p id=\"fs-id1170572135837\">Notice that we did not include the \u201c+ <em>C<\/em>\u201d term when we wrote the antiderivative. The reason is that, according to the Fundamental Theorem of Calculus, Part 2, <em>any<\/em> antiderivative works. So, for convenience, we chose the antiderivative with [latex]C=0.[\/latex] If we had chosen another antiderivative, the constant term would have canceled out. This always happens when evaluating a definite integral.<\/p>\n<p id=\"fs-id1170571660070\">The region of the area we just calculated is depicted in Figure 3. Note that the region between the curve and the [latex]x[\/latex]-axis is all below the [latex]x[\/latex]-axis. Area is always positive, but a definite integral can still produce a negative number (a net signed area). For example, if this were a profit function, a negative number indicates the company is operating at a loss over the given interval.<\/p>\n<div style=\"width: 335px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11204110\/CNX_Calc_Figure_05_03_004.jpg\" alt=\"The graph of the parabola f(t) = t^2 \u2013 4 over &#091;-4, 4&#093;. The area above the curve and below the x axis over &#091;-2, 2&#093; is shaded.\" width=\"325\" height=\"283\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3. The evaluation of a definite integral can produce a negative value, even though area is always positive.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571561288\" class=\"textbook exercises\">\n<h3>Example: Evaluating a Definite Integral Using the Fundamental Theorem of Calculus, Part 2<\/h3>\n<p id=\"fs-id1170572607951\">Evaluate the following integral using the Fundamental Theorem of Calculus, Part 2:<\/p>\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{1}^{9}\\dfrac{x-1}{\\sqrt{x}}dx.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572130389\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572130389\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572130389\">First, eliminate the radical by rewriting the integral using rational exponents. Then, separate the numerator terms by writing each one over the denominator:<\/p>\n<div id=\"fs-id1170572130394\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{1}^{9}\\frac{x-1}{{x}^{1\\text{\/}2}}dx={\\displaystyle\\int }_{1}^{9}(\\frac{x}{{x}^{1\\text{\/}2}}-\\frac{1}{{x}^{1\\text{\/}2}})dx\\text{.}[\/latex]<\/div>\n<p id=\"fs-id1170572624682\">Use the properties of exponents to simplify:<\/p>\n<div id=\"fs-id1170572233529\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{1}^{9}(\\frac{x}{{x}^{1\\text{\/}2}}-\\frac{1}{{x}^{1\\text{\/}2}})dx={\\displaystyle\\int }_{1}^{9}({x}^{1\\text{\/}2}-{x}^{-1\\text{\/}2})dx\\text{.}[\/latex]<\/div>\n<p id=\"fs-id1170572563042\">Now, integrate using the power rule:<\/p>\n<div id=\"fs-id1170572563045\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\\\ {\\displaystyle\\int }_{1}^{9}({x}^{1\\text{\/}2}-{x}^{-1\\text{\/}2})dx\\hfill & ={(\\frac{{x}^{3\\text{\/}2}}{\\frac{3}{2}}-\\frac{{x}^{1\\text{\/}2}}{\\frac{1}{2}})|}_{1}^{9}\\hfill \\\\ \\\\ & =\\left[\\frac{{(9)}^{3\\text{\/}2}}{\\frac{3}{2}}-\\frac{{(9)}^{1\\text{\/}2}}{\\frac{1}{2}}\\right]-\\left[\\frac{{(1)}^{3\\text{\/}2}}{\\frac{3}{2}}-\\frac{{(1)}^{1\\text{\/}2}}{\\frac{1}{2}}\\right]\\hfill \\\\ & =\\left[\\frac{2}{3}(27)-2(3)\\right]-\\left[\\frac{2}{3}(1)-2(1)\\right]\\hfill \\\\ & =18-6-\\frac{2}{3}+2\\hfill \\\\ & =\\frac{40}{3}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170572522399\">See Figure 4.<\/p>\n<div style=\"width: 335px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11204112\/CNX_Calc_Figure_05_03_005.jpg\" alt=\"The graph of the function f(x) = (x-1) \/ sqrt(x) over &#091;0,9&#093;. The area under the graph over &#091;1,9&#093; is shaded.\" width=\"325\" height=\"246\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 4. The area under the curve from [latex]x=1[\/latex] to [latex]x=9[\/latex] can be calculated by evaluating a definite integral.<\/p>\n<\/div>\n<div class=\"wp-caption-text\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Evaluating a Definite Integral Using the Fundamental Theorem of Calculus, Part 2.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/UdsTNaiWmbs?controls=0&amp;start=998&amp;end=1063&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266833\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266833\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/5.3TheFundamentalTheoremOfCalculus998to1063_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.3 The Fundamental Theorem of Calculus&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1170571609442\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Use the second part of the Fundamental Theorem of Calculus to evaluate [latex]{\\displaystyle\\int }_{1}^{2}{x}^{-4}dx.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q3887125\">Hint<\/span><\/p>\n<div id=\"q3887125\" class=\"hidden-answer\" style=\"display: none\">\n<p>Use the power rule.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571639103\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571639103\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{7}{24}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm211336\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=211336&theme=oea&iframe_resize_id=ohm211336&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div id=\"fs-id1170571597325\" class=\"textbook exercises\">\n<h3>Example: A Roller-Skating Race<\/h3>\n<p>James and Kathy are racing on roller skates. They race along a long, straight track, and whoever has gone the farthest after 5 sec wins a prize. If James can skate at a velocity of [latex]f(t)=5+2t[\/latex] ft\/sec and Kathy can skate at a velocity of [latex]g(t)=10+ \\cos (\\frac{\\pi }{2}t)[\/latex] ft\/sec, who is going to win the race?<\/p>\n<div id=\"fs-id1170571597327\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572332613\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572332613\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572332613\">We need to integrate both functions over the interval [latex]\\left[0,5\\right][\/latex] and see which value is bigger. For James, we want to calculate<\/p>\n<div id=\"fs-id1170572332632\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{0}^{5}(5+2t)dt.[\/latex]<\/div>\n<p id=\"fs-id1170571758991\">Using the power rule, we have<\/p>\n<div id=\"fs-id1170571758994\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int }_{0}^{5}(5+2t)dt\\hfill & ={(5t+{t}^{2})|}_{0}^{5}\\hfill \\\\ & =(25+25)=50.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170571772586\">Thus, James has skated 50 ft after 5 sec. Turning now to Kathy, we want to calculate<\/p>\n<div id=\"fs-id1170571772589\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\displaystyle\\int }_{0}^{5}10+ \\cos (\\frac{\\pi }{2}t)dt.[\/latex]<\/div>\n<p id=\"fs-id1170572292292\">We know [latex]\\sin t[\/latex] is an antiderivative of [latex]\\cos t,[\/latex] so it is reasonable to expect that an antiderivative of [latex]\\cos (\\frac{\\pi }{2}t)[\/latex] would involve [latex]\\sin (\\frac{\\pi }{2}t).[\/latex] However, when we differentiate [latex]\\sin (\\frac{\\pi }{2}t),[\/latex] we get [latex]\\frac{\\pi }{2} \\cos (\\frac{\\pi }{2}t)[\/latex] as a result of the chain rule, so we have to account for this additional coefficient when we integrate. We obtain<\/p>\n<div id=\"fs-id1170571599308\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int }_{0}^{5}10+ \\cos (\\frac{\\pi }{2}t)dt\\hfill & ={(10t+\\frac{2}{\\pi } \\sin (\\frac{\\pi }{2}t))|}_{0}^{5}\\hfill \\\\ & =(50+\\frac{2}{\\pi })-(0-\\frac{2}{\\pi } \\sin 0)\\hfill \\\\ & \\approx 50.6.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170572368405\">Kathy has skated approximately 50.6 ft after 5 sec. Kathy wins, but not by much!<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: A Roller-Skating Race.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/UdsTNaiWmbs?controls=0&amp;start=1072&amp;end=1258&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266835\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266835\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/5.3TheFundamentalTheoremOfCalculus1072to1258_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.3 The Fundamental Theorem of Calculus&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1170572368412\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1170572368419\">Suppose James and Kathy have a rematch, but this time the official stops the contest after only 3 sec. Does this change the outcome?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170571773427\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170571773427\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571773427\">Kathy still wins, but by a much larger margin: James skates 24 ft in 3 sec, but Kathy skates 29.3634 ft in 3 sec.<\/p>\n<h4>Hint<\/h4>\n<p id=\"fs-id1170572368423\">Change the limits of integration from those in the previous example.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571773434\" class=\"textbox tryit\">\n<h3>Activity: A Parachutist in Free Fall<\/h3>\n<p>&nbsp;<\/p>\n<div style=\"width: 711px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11204117\/CNX_Calc_Figure_05_03_006.jpg\" alt=\"Two skydivers free falling in the sky.\" width=\"701\" height=\"462\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 5. Skydivers can adjust the velocity of their dive by changing the position of their body during the free fall. (credit: Jeremy T. Lock)<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572218426\">Julie is an avid <span class=\"no-emphasis\">skydiver<\/span>. She has more than 300 jumps under her belt and has mastered the art of making adjustments to her body position in the air to control how fast she falls. If she arches her back and points her belly toward the ground, she reaches a terminal velocity of approximately 120 mph (176 ft\/sec). If, instead, she orients her body with her head straight down, she falls faster, reaching a terminal velocity of 150 mph (220 ft\/sec).<\/p>\n<p id=\"fs-id1170572218437\">Since Julie will be moving (falling) in a downward direction, we assume the downward direction is positive to simplify our calculations. Julie executes her jumps from an altitude of 12,500 ft. After she exits the aircraft, she immediately starts falling at a velocity given by [latex]v(t)=32t.[\/latex] She continues to accelerate according to this velocity function until she reaches terminal velocity. After she reaches terminal velocity, her speed remains constant until she pulls her ripcord and slows down to land.<\/p>\n<p id=\"fs-id1170572419103\">On her first jump of the day, Julie orients herself in the slower \u201cbelly down\u201d position (terminal velocity is 176 ft\/sec). Using this information, answer the following questions.<\/p>\n<ol id=\"fs-id1170572419110\">\n<li>How long after she exits the aircraft does Julie reach terminal velocity?<\/li>\n<li>Based on your answer to question 1, set up an expression involving one or more integrals that represents the distance Julie falls after 30 sec.<\/li>\n<li>If Julie pulls her ripcord at an altitude of 3000 ft, how long does she spend in a free fall?<\/li>\n<li>Julie pulls her ripcord at 3000 ft. It takes 5 sec for her parachute to open completely and for her to slow down, during which time she falls another 400 ft. After her canopy is fully open, her speed is reduced to 16 ft\/sec. Find the total time Julie spends in the air, from the time she leaves the airplane until the time her feet touch the ground.<br \/>\nOn Julie\u2019s second jump of the day, she decides she wants to fall a little faster and orients herself in the \u201chead down\u201d position. Her terminal velocity in this position is 220 ft\/sec. Answer these questions based on this velocity:<\/li>\n<li>How long does it take Julie to reach terminal velocity in this case?<\/li>\n<li>Before pulling her ripcord, Julie reorients her body in the \u201cbelly down\u201d position so she is not moving quite as fast when her parachute opens. If she begins this maneuver at an altitude of 4000 ft, how long does she spend in a free fall before beginning the reorientation?<br \/>\nSome jumpers wear \u201c<span class=\"no-emphasis\">wingsuits<\/span>\u201d (see Figure 6). These suits have fabric panels between the arms and legs and allow the wearer to glide around in a free fall, much like a flying squirrel. (Indeed, the suits are sometimes called \u201cflying squirrel suits.\u201d) When wearing these suits, terminal velocity can be reduced to about 30 mph (44 ft\/sec), allowing the wearers a much longer time in the air. Wingsuit flyers still use parachutes to land; although the vertical velocities are within the margin of safety, horizontal velocities can exceed 70 mph, much too fast to land safely.<\/p>\n<div style=\"width: 710px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11204124\/CNX_Calc_Figure_05_03_007.jpg\" alt=\"A person falling in a wingsuit, which works to reduce the vertical velocity of a skydiver\u2019s fall.\" width=\"700\" height=\"392\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 6. The fabric panels on the arms and legs of a wingsuit work to reduce the vertical velocity of a skydiver\u2019s fall. (credit: Richard Schneider)<\/p>\n<\/div>\n<\/li>\n<\/ol>\n<p id=\"fs-id1163722683509\">Answer the following question based on the velocity in a wingsuit.<\/p>\n<ol id=\"fs-id1163722683512\" start=\"7\">\n<li>If Julie dons a wingsuit before her third jump of the day, and she pulls her ripcord at an altitude of 3000 ft, how long does she get to spend gliding around in the air?<\/li>\n<\/ol>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-546\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>5.3 The Fundamental Theorem of Calculus. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 1. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\">https:\/\/openstax.org\/details\/books\/calculus-volume-1<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":15,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"5.3 The Fundamental Theorem of Calculus\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-546","chapter","type-chapter","status-publish","hentry"],"part":57,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/546","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":19,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/546\/revisions"}],"predecessor-version":[{"id":4877,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/546\/revisions\/4877"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/57"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/546\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=546"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=546"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=546"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=546"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}